A van moves with a constant speed of 20 miles per hour. How far can it travel in 3 1/2 hours

A cyclist reaches the top of a hill with

rjkz [21]

Answer:Based on the excerpt, which best describes Harburg’s view of the Great Depression?

He has no interest in financial success for himself.

He values artistic success over financial success for himself.

He believes most people benefited from losing their financial stability.

He regrets the fact that he gave away his money to benefit his art.

Explanation:

5 0

2 years ago

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

3 years ago

an 1150kg elevator moving down speeds up at a rate of 3.5m/s. what is the tension in the supporting cables?

gtnhenbr [62]

Answer:

The tension force in the supporting cables is 7245N

Explanation:

There are two forces acting on the elevator: the force of gravity pointing down (+) with magnitude (elevator mass) x (gravitational acceleration), and the tension force of the cable pointing up (-) with an unknown magnitude F. The net force is the sum of these forces:

$F_{net} = F_g - F = m\cdot g - F\\$

We are given the resulting acceleration along with the mass, i.e., we know the net force, allowing us to solve for F:

$1150kg\cdot 3.5\frac{m}{s^2}= 1150kg \cdot 9.8\frac{m}{s^2}-F\\\implies F = 1150kg\cdot(9.8-3.5)\frac{m}{s^2}= 7245N$

The tension force F in the supporting cables is 7245N

3 0

3 years ago

A motorcycle accelerates uniformly from rest at 7.9\,\dfrac{\text{m}}{\text{s}^2}7.9 s 2 m 7, point, 9, space, start fraction,

8090 [49]

Answer:

t = 3.516 s

Explanation:

The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:

$v(t) = v_0 +at$

Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0

If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.

This can be done knowing that

1 km = 1000 m

1 h = 3600 s

Therefore

1 km/h = (1000/3600) m/s = 0.2777... m/s

100 km/h = 27.777... m/s

Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:

27.777 m/s = 7.9 m/s^2 t

Solving for t

t = (27.7777 / 7.9) s = 3.516 s

6 0

3 years ago

At a baseball game, Bill observes from the bleachers Jack throwing a ball toward home at 15 km/s while Kevin runs toward home fr

sattari [20]

Answer:bill 5 m/s. Jack:10 m/s

Explanation:

Cuz I took it

7 0

2 years ago